Note that by Lagrange's theorem, the order of any subgroup must divide the order of the group. Thus, the order of any proper nontrivial subgroup is one of the numbers 2,4,8,3,6,12,24,5,10,20,40,15,30,60.

Case : The number of subgroups of order 8 divides . Combining with the congruence condition, we get that the number of subgroups of order 8 is 1, 3, 5, or 15.

Case : The number of subgroups of order 3 divides . Combining with the congruence condition, we get that the number of subgroups of order 3 is 1 or 4.

Case : The number of subgroups of order 5 divides . Combining with the congruence condition, we get that the number of subgroups of order 5 is 1 or 6.

In the case of a finite nilpotent group, the number of subgroups of a given order is the product of the number of subgroups of order equal to each of its maximal prime power divisors, in the corresponding Sylow subgroup. In particular, we get (number of subgroups of order 8) = (number of subgroups of order 3) = (number of subgroups of order 5) = (number of subgroups of order 24) = (number of subgroups of order 15) = (number of subgroups of order 40) = 1, and also, (number of subgroups of order 2) = (number of subgroups of order 6) = (number of subgroups of order 10) = (number of subgroups of order 30). Separately, we have (number of subgroups of order 4) = (number of subgroups of order 12) = (number of subgroups of order 20) = (number of subgroups of order 60).

Subgroup lattice and quotient lattice of finite abelian group are isomorphic: This states that the number of subgroups of order equals the number of subgroups of order . In particular, we get (number of subgroups of order 8) = (number of subgroups of order 3) = (number of subgroups of order 5) = (number of subgroups of order 24) = (number of subgroups of order 15) = (number of subgroups of order 40) = 1. Similarly, we have (number of subgroups of order 2) = (number of subgroups of order 4) = (number of subgroups of order 6) = (number of subgroups of order 12) = (number of subgroups of order 10) = (number of subgroups of order 20) = (number of subgroups of order 30) = (number of subgroups of order 60).